Develop Reference

How to get a user-selected file and feed it to a python program in Tkinter

I am currently messing about with Tkinter creating an interface for a project I created. The program takes in a bunch of file paths as inputs for it to run. I'm trying to create a tkinter interface where I can upload the 4 files I need or at least somehow get the filepaths and the. feed those to the program. Here is what I have:
import sys
import os
import comparatorclass
from tkinter import *
from tkinter.ttk import *
from tkinter.filedialog import askopenfile
root=Tk()
root.geometry('1000x1000')
def open_file():
file = askopenfile(mode ='r', filetypes =[('Python Files', '*.py')])
if file is not None:
content = file.read()
print(content)
def run_comparator():
comparatorclass.main()
button2 = Button(root, text ='Open', command = lambda:open_file())
button2.pack(side = TOP, pady = 10)
button1 = Button(root,text="hello",command= run_comparator)
button1.pack()
root.mainloop()
as you can see, I have two buttons. The issue I'm having is how to connect my openfile function to my run_comparator function such that the 4 files I need to open are passed on to the run_comparator

Getting folder path with Python Tkinter tkFileDialog

Using Python Tkinter I am trying to get the directory path of selected Folder. I do not want to load a file or navigate to a file but get the folder path like
How can I do this?
from Tkinter import *
from tkFileDialog import askopenfilename
def callback():
name= askopenfilename()
print name
errmsg = 'Error!'
Button(text='File Open', command=callback).pack(fill=X)
mainloop()
Update
from Tkinter import *
from tkFileDialog import askopenfilename
from tkinter import filedialog #for Python 3
def callback():
name= askopenfilename()
directory = filedialog.askdirectory()
print directory
errmsg = 'Error!'
Button(text='File Open', command=callback).pack(fill=X)
mainloop()

You can use askdirectory from filedialog as follows:
from tkinter import filedialog #for Python 3
directory = filedialog.askdirectory()

Ok Looks like I find the solution on my own. Putting here which might help someone else in future.
import Tkinter, tkFileDialog
root = Tkinter.Tk()
root.withdraw()
dirname = tkFileDialog.askdirectory(parent=root,initialdir="/",title='Please select a directory')
print(dirname)

Using tkinter to select and save a file

I am looking to create a reminder that prompts the user to select a file if they've completed it. After selecting the file they want, I would like the script to save that file to a specific folder. This is what I have so far...
import tkinter
import time
import win32api
from tkinter import messagebox
dialog_title = 'Complaince calendar'
dialog_text = 'Have you completed the monthly spreadsheet?'
answer = messagebox.askquestion(dialog_title, dialog_text)
if answer == 'yes':
import tkinter as tk
from tkinter import filedialog
root = tk.Tk()
root.withdraw()
file_path = filedialog.askopenfilename()
else: # 'no'
win32api.MessageBox(0, 'Make sure you get that done', 'Compliance calendar')
How can I save the file_path to a specified folder?
Thanks!!

How to open a exe file using python Tkinter

I am trying to open a exe file from C drive by clicking a button from my GUI and up till now i am unable to select the specific file. Can i know if there is any function to directly open the file in Tkinter. Currently,tkFileDialog.askdirectory only direct me to the FILES.
import Tkinter
import tkMessageBox
import tkFileDialog
import os
import subprocess
top = Tkinter.Tk()
def run():
File = tkFileDialog.askdirectory()
os.system(File)
b = Tkinter.Button(top, text = 'DAQoutput', command= run)
b.pack()
top.mainloop()

Directly using Tkinter? No there is no command for it but you don't need one. Tkinter is a GUI toolkit and that's all it does.
Instead use os.system() to launch your exe.
Your script should look something like this:
import Tkinter
import tkMessageBox
import tkFileDialog
import os
top = Tkinter.Tk()
def run(self):
File = tkFileDialog.askdirectory()
os.system(File)
b = Tkinter.Button(top, text = 'DAQoutput', command= self.run)
b.pack()
top.mainloop()

Open File Dialog freezes after tkinter askopenfilename method is called

I'm trying to simply get a file name from the user by tkinter.filedialog.askopenfilename(). The function returns fine and the code below displays the file name okay but the dialog window doesn't close immediately after hitting 'open' or 'cancel', it freezes. I'm using python 3.3.3 or OSX 10.9.1 and tcl/tK 8.5.9.
from tkinter import *
from tkinter.messagebox import *
from tkinter.filedialog import *
top = Tk()
top.withdraw()
file_name = filedialog.askopenfilename()
print (file_name)

Adding root.update() after filedialog.askopenfilename() gets the open file dialog to close after the file is chosen.
root = tk.Tk()
root.withdraw()
file_path = filedialog.askopenfilename()
root.update()
Refer to: Tkinter askopenfilename() won't close

A work around that I used for this is to "withdraw" the tkinter window AFTER selecting the file from file explorer. Here is the code snippet I used -
import tkinter
from tkinter import filedialog
def selectCustomerFileTK():
root = tkinter.Tk()
root.wm_attributes('-topmost', 1)
filename = filedialog.askopenfilename()
root.withdraw()
return filename
getfile = selectCustomerFileTK()
It opens a tkinter window while you are selecting the file, but the moment you select the file and press "Open", the tkinter window and file explorer both close because of the "root.withdraw()" command after.

You don't need to specify module name in
file_name = filedialog.askopenfilename()
Try
file_name = askopenfilename()
instead